Gspan: Graph-based Substructure Pattern Mining

Presented By: Sadik MussahUniversity of Vermont CS 332 – Data mining

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- Algorithm -

Outlines• Background• Problem Definition• Authors Contribution• Concepts Behind Gspan• Experimental Result• Conclusion

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Background• Frequent Subgraph Mining Is An Extension To

Existing Frequent Pattern Mining Algorithms• A Major Challenge Is To Count How Many

Instances of patterns are in the Dataset• Counting Instances Might Be Easy For Sets, But

Subtle For Graphs• Graph Isomorphism Problem

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Background

Theorem

Given two graphs G and G’ (g prime), G isomorphic to G’ iff min(G) = min(G’)

05/01/23Sadik Mussah

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Background5

X W

U Y

V

(a)X

W

U

YV

(b)Two Isomorphic graph (a) and (b) with their mapping

function (c) Two Graphs Are Isomorphic If One Can Find A Mapping Of

Nodes Of The First Graph To The Second Graph Such That Labels On Nodes And Edges Are Preserved.

f(V1.1) = V2.2f(V1.2) = V2.5f(V1.3) = V2.3f(V1.4) = V2.4f(V1.5) = V2.1

(c)

G1=(V1,E1,L1) G2=(V2,E2,L2)

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Problem: Finding Frequent Subgraphs

• Problem Setting: Similar To Finding Frequent Itemsets For Association Rule Discovery • Input: Database Of Graph Transactions

• Undirected Simple Graph (No Multiples Edges)• Each Graph Transaction Has Labeled Edges/Vertices.• Transactions May Not Be Connected • Minimum Support Thresholds

• Output: Frequent Subgraphs That Satisfy The Support Threshold, Where Each Frequent Subgraph Is Connected.

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Finding Frequent Subgraphs7

Authors Contribution• Representing Graphs As Strings (Like Treeminer)• No Candidate Generation!• “It Combines The Growing And Checking Of Frequent

Subgraphs Into One Procedure, Thus Accelerates The Mining Process.”• Really Fast, Still A Standard Baseline System That

Most Rivals Compare Their Systems To.

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Concepts Behind Gspan• The Idea Is To Produces A Depth-first Search (DFS) Codes

For Each Edge In Graphs• Edges Are Sorted According To Lexicographic Order Of

Codes• Yan And Han Proved That Graph Isomororphism Can Be

Tested For Two Graphs Annotated With DFS Codes• Starting With Small Graph Patterns Containing 1-edge,

Patterns Are Expanded Systemically By The DFS Search

• Employ Anti-monotonic Property Of Graph Frequency

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Lexicographic Ordering In Graph

• It Can Tell Us The Order Of Two Graphs.• The Design Can Help Us Build A Similar Hierarchy.• The Design Should Guarantee Easy-growing From One

Level To The Lower Level And Easy-rolling-up From Low Level To Higher Level.• It May Be Difficult To Have Such Design That No Two

Nodes In This Tree Are Same For Graph Case. • It Can Tell Us Whether The Graph Has Been

Discovered.• And More, The Most Important, If A Graph Has Been

Discovered, All Its Children Nodes In The Hierarchy Must Have Been Discovered.

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Lexicographic Ordering in Graph

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...

... ...

1-edge

2-edge

...3-edge ...

......

...

DFS Code And Minimum DFS Code• We Use A 5-tuple (Vi, Vj, L(vi), L(vj), L(vi,vj)) To Represent An

Edge. (It May Be Redudant, But Much Easier To Understand.)• Turn A Graph Into A Sequence Whose Basic Element Is 5-tuple.

Form The Sequence In Such An Order:• To Extend One New Node, Add The Forward

Edge That Connect One Node In The Old Graph With This New Node.• Add All Backward Edge That Connect This

New Node To Other Nodes In The Old Graph• Repeat This Procedure.

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DFS code13

XY

X

Z

Z

a ab

b c

d

v0v1v2

v3v4

XY

ae0: (0,1,x,y,a)

Xb

e1: (1,2,y,x,b)a

e2: (2,0,x,x,a)

Zc e3: (2,3,x,z,c)b

e4: (3,1,x,y,b)

Zd

e5: (1,4,x,z,d)

DFS Code And Minimum DFS Code14

Depth First Tree And Forward/Backward Edge Set

Minimum DFS code15

Each Graph may have lots of DFS code (why?):one smallest lexicographic one is its Minimum DFS CodeEdge no. (B) (C) (D)

0 (0,1,x,y,a) (0,1,y,x,a) (0,1,x,x,a)

1 (1,2,y,x,b) (1,2,x,x,a) (1,2,x,y,b)

2 (2,0,x,x,a) (2,0,x,y,b) (0,1,y,x,a)

3 (2,3,x,z,c) (2,3,x,z,c) (2,3,y,z,a)

4 (3,1,z,y,b) (3,0,z,y,b) (3,1,z,x,c)

5 (1,4,x,z,d) (0,4,y,z,d) (2,4,y,z,d)

Graph Parent And Its Children16

X

Y

X

Z Z

a

b

ca

Given a DFS code c0=(e0,e1,…,en)if c1=(e0,e1,…,en,ex)if c0<c1, then c0 is c1’s parent,c1 is c0’s child.

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Theorem

• 1. Given Two Graph G0 And G1, G0 Is Isomorphic To G1 Iff Min_dfs_code(g0)=min_dfs_code(g1).

• 2. DFS Code Tree Covers All Graphs Although Some Tree Nodes May Represent The Same Graph

• 3. Given A Node In DFS Code Tree, If Its DFS Code Is Not Its Minimum DFS Code, Prune This Node And Its All Descendants Won’t Change. “Covering”.

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DFS Code Tree18

...

... ...

1-edge

2-edge

...3-edge ...

......

...

pruned

FSG: two substructure patterns and their potential candidates.

05/01/23Sadik Mussah

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05/01/23SADIK MUSSAH

20AGM: two substructures joined by two chains

Algorithm21

Algorithm22

Algorithm: Apriorigraph

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ALGORITHM: gSpan

05/01/23Xifeng Yan

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Experimental Result25

Experimental Result26

Conclusion• No Candidate Generation And False Test• Space Saving From Depth First Search • Good Performance: Using “Memory Pool” And

One Major Counting Improvement, It Seems The Performance Will Be Improved 5 Times More. (But Need More Testing).

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QuestionsQ1) What Two Major Costs From Apriori-like,

Frequent Substructure Mining Algorithms Did Gspan Aim To Reduce/Avoid?

Answer:1) The Creation Of Size K+1 Candidate Subgraphs From Size K Frequent Subgraphs Is More Complicated And Costly The Standard Apriori Large Itemset Generation. 2) Pruning False Positives Is An Expensive Process. Subgraph Isomorphism Problem Is Np-complete.

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Security Graph 3D Visualization

• https://www.youtube.com/watch?v=JsEm-CDj4qM

05/01/23Sadik Mussah

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Questions (cont.)• Q2) Which DFS Tree Does The DFS Code Below Belong To?

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v0Y

x

x

z

z v4v1

v2

v3

a

a

c

bb

d

Answer: tree (c)

Questions• Q3) What Does Gspan Compare When Testing For

Isomorphism Between Two Graphs, And Why?

• Answer: Gspan Compares The Minimum Dfs Codes Of The Two Graphs. Given Two Graphs G And G’, G Is Isomorphic To G’ If Min(g)=min(g’). This Theorem Allows For A Simple String Comparison Of More Complicated Graphs. If Two Nodes Contain The Same Graph But Different Minimum DFS Codes, We Can Prune The Sub-branch Of The Rightmost Of The Two Nodes. This Greatly Decreases The Problem Size.

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Questions?33